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The Boltzmann distribution assigns probabilities to individual microstates in a canonical ensemble, a statistical ensemble for thermal contact. The system exchanges energy with a thermal reservoir at fixed positive absolute temperature $T$, while its particle number $N$ and external parameters, such as volume, stay fixed. The thermal reservoir is large enough that these exchanges barely change its temperature; the coupling permits thermal equilibrium but contributes negligible interaction energy. All the microstates considered have the same $N$. No chemical potential term is needed in their weights; fixed $N$ does not imply that the chemical potential is zero.

For a finite discrete set of microstates with energies $E_i$, writeHere $k_B$ is the Boltzmann constant, $\beta$ is the inverse temperature, $e^{-\beta E_i}$ is a Boltzmann factor, and $Z$ is the partition function. The Boltzmann factor is a weight; dividing by the partition function makes it a probability. ^[ref]

The useful surprise is that a fixed temperature still permits thermal fluctuations in the system's energy. The Boltzmann distribution predicts their probabilities and expected values without tracking every microscopic exchange. Its exponential function also makes rare excited microstates sensitive to temperature: an energy gap much larger than $k_{B}T$ gives strong Boltzmann suppression, while warming increases their relative populations. A population ratio can therefore reveal a gap or serve as a thermometer, provided thermal equilibrium and the relevant state degeneracies are known.

Historically, Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) presents the canonical ensemble in Chapter IV. His equations 90–92 give an exponential function of energy and its normalization, in classical phase space. This classical formulation exhibits the same weighting principle. ^{[ref] [ref]}

Group microstates by their energy $E$. The state degeneracy $g(E)$ counts how many microstates have that energy; their aggregate probability isHere $P(E)$ is the probability of the macrostate retaining only energy, while $p_i$ is the probability of one individual microstate. For two such macrostates,Counting more microstates can compensate for a smaller Boltzmann factor. Omitting the state degeneracy confuses a sum over microstates with a sum over distinct energies. ^[ref]

Consider two distinguishable, noninteracting two-level systems, each with energies $0$ and $\Delta >0$. Their four microstates $(0,0),(0,1),(1,0),(1,1)$ have total energies $0,\Delta ,\Delta ,2\Delta$. Thus the three energy macrostates have state degeneracies $1,2,1$. This is a stipulated finite model, not a measured material.

Set $q=e^{-\beta \Delta }$. The four individual Boltzmann factors are $1,q,q,q^2$, so the partition function and aggregate probabilities areEach of the two middle microstates has probability $q/Z$, while their shared energy macrostate has probability $2q/Z$.

At the temperature satisfying $\Delta =k_{B}T\ln 2$, the Boltzmann factor is $q=1/2$ and the partition function is $Z=9/4$. ThereforeThe middle energy is as likely as the ground state energy because its state degeneracy is twice as large. Yet each individual middle microstate is only half as likely as the ground state microstate. The diagram places the four energies beside their individual probabilities, then shows the sums within each macrostate.

The expected value of energy follows directly by weighting the three possible energies:At $q=1/2$ this is $2\Delta /3$. An individual energy measurement still returns $0$, $\Delta$ or $2\Delta$; the expected value describes the ensemble average, rather than a fourth possible measured energy.

Keep the gap $\Delta$ fixed and vary positive temperature. With $x=k_{B}T/\Delta$, the Boltzmann factor is $q=e^{-1/x}$. In the four-microstate example, as $T\to 0^{+}$, $q\to 0$: the unique ground state has probability tending to one, and the other microstates have probabilities tending to zero. If several microstates instead shared the lowest energy, their state degeneracy would split that limiting population equally.

As $T\to \infty$ in this fixed finite model, all four individual Boltzmann factors tend to one, so all four individual microstates have probability tending to $1/4$. The three aggregate probabilities instead approachThe equal-population limit applies to this finite set of microstates. An infinite collection of microstates cannot all have the same positive normalized probability, so the statement cannot be extended to an unbounded energy spectrum. Real systems may also acquire additional accessible microstates as temperature rises.

The plotted crossing occurs at $x=1/\ln 2$, where $P(0)=P(\Delta )=4/9$. For larger positive temperatures, the middle macrostate is more probable than the lowest-energy macrostate, although the lowest-energy individual microstate remains more probable than either individual middle microstate. The expected value of energy rises from zero toward $\Delta$ in this finite model.

The partition function $Z(\beta )={\sum }_s{e}^{-\beta E_s}$ is the sum of the Boltzmann weights over individual microstates. Dividing by it makes their probabilities sum to one. When summing distinct energies instead, include the state degeneracy: $Z={\sum }_{E}g(E)e^{-\beta E}$. For two independent distinguishable two-level systems,The middle coefficient counts two microstates, exactly as in the preceding figure. For $N$ such units, $Z_N=Z_1^N$; this factorization requires additive energy and independence, and generally fails with interactions.

The usefulness goes beyond normalization. For fixed energies, derivatives of the natural logarithm of the partition function yield measurable quantities:Here $U$ is internal energy, the second expression is the energy variance, and $S$ is Gibbs entropy. For one two-level system at fixed $\Delta$, its heat capacity $C=\partial U/\partial T$ and entropy areAt $\beta \Delta =\ln 3$, these give $U/\Delta =0.25$, $C/k_B\simeq 0.2263$ and $S/k_B\simeq 0.5623$. For independent units, all three quantities add.

The figure shows why heat capacity peaks. At low temperature, the excited energy is rarely occupied; at high temperature, the probabilities are almost equal and further warming barely changes internal energy. The largest response lies between these limits, at $k_{B}T/\Delta \simeq 0.4168$, where $C/k_B\simeq 0.4392$. Meanwhile entropy rises from zero toward $k_B\ln 2$, reflecting two equally likely alternatives. This high-temperature limit belongs to the stipulated two-level system; extra accessible energy eigenstates would change it.

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

Take a cycle graph with four labeled graph vertices and four distinct edges $\{1,2\},\{2,3\},\{3,4\},\{4,1\}$. Set $h=0$, $J>0$, and impose periodic boundary conditions by writing ${\sigma }_5={\sigma }_1$. The square drawing represents a one-dimensional ring: it has no diagonal interactions and is not a two-dimensional lattice.

Call an edge between opposite Ising spins a domain wall. If $d$ is their number, the four bond products sum to $4-2d$, so the energy isEach domain wall costs $2J$ relative to an aligned bond. Returning around the ring must return to the starting Ising spin, so $d$ is even: equivalently, ${\prod }_{i=1}^4({\sigma }_i{\sigma }_{i+1})=1$. Choosing an even set of $d$ wall edges and then choosing ${\sigma }_1$ determines the entire pattern. The state degeneracy is therefore $g_d=2\left(\genfrac{}{}{0ex}{}{4}{d}\right)$, using a binomial coefficient.

The twelve middle microstates consist of eight patterns with one exceptional Ising spin and four patterns with two adjacent $+1$ Ising spins and two adjacent $-1$ Ising spins. The two alternating patterns have four domain walls. All $2+12+2=16$ microstates are shown below.

Writing $K=\beta J$, the partition function and the probability of each energy class areFor an original exact example, choose $K=(\ln 2)/4$. Then $e^{4K}=2$, $Z=17$, and $P(d=0)=4/17$, $P(d=2)=12/17$, $P(d=4)=1/17$. Each aligned microstate has probability $2/17$, each middle microstate $1/17$, and each alternating microstate $1/34$: multiply by state degeneracy before comparing whole classes. As $K\to 0$, all sixteen microstates become equally likely, but the class probabilities approach $1/8,3/4,1/8$. As $K\to \infty$, the two all-aligned microstates each have limiting probability $1/2$.

A macrostate groups microstates sharing prescribed coarse data. The grouping depends on what is retained: here retain only total energy. For the two distinguishable two-level systems, the macrostates $E=0,\Delta ,2\Delta$ have state degeneracies $1,2,1$. Although two units are not macroscopic in size, they make this loss of information visible: knowing $E=\Delta$ does not tell us which unit is excited.

At inverse temperature $\beta \Delta =\ln 3$, each unit's excited-state probability is $p=1/4$. Their independence gives microstate probabilities $9/16,3/16,3/16,1/16$ in the order above. Adding within each macrostate givesThe total expected value of energy is therefore $\Delta (6/16)+2\Delta (1/16)=\Delta /2$, matching two units each with expected value of energy $\Delta /4$. Equal state degeneracies would give the wrong answer: macrostate probabilities combine counting with the Boltzmann factor. At very high positive temperature, all four microstates become equally likely, but these three macrostate probabilities approach $1/4,1/2,1/4$.

A uniform magnetic field quantizes the transverse cyclotron motion of an electron. With electron charge $-e$ and the Landau gauge for a uniform magnetic field $\mathbf{A}=(0,Bx,0)$, minimal coupling givesFor an eigenstate of $p_y$ with eigenvalue $\hslash k_y$, this is a quantum harmonic oscillator centered at $x_c=-\hslash k_y/(eB)$, with cyclotron frequency ${\omega }_c=eB/m$. Its Landau levels areThe oscillator width is set by the magnetic length ${\ell }_B=\sqrt{\hslash /(eB)}$. In three dimensions, the free motion parallel to the magnetic field adds ${\hslash }^2{k}_z^2/(2m)$. The electron spin adds a Zeeman splitting $\pm g_s{\mu }_{B}B/2$. In a crystal the appropriate band effective mass and effective spin factor replace the free-electron values.

Changing $k_y$ moves the orbit center without changing the energy. In a rectangle $L_x{L}_y$, adjacent allowed centers differ by $2\pi {\ell }_B^2/L_y$. Requiring the center to lie in the sample gives the degeneracy of a Landau level, per spin,This large state degeneracy, together with the increasing level spacing, rearranges the occupied states as the magnetic field changes. To resolve the Landau levels, thermal broadening $k_{B}T$ and disorder broadening should both be small compared with $\hslash {\omega }_c$.

The de Haas-van Alphen effect is an oscillation of magnetization as successive Landau levels pass the Fermi energy. The Shubnikov-de Haas effect is the corresponding oscillation in longitudinal magnetoresistance. For a simple orbit, successive crossings occur periodically in $1/B$; the general relation to the extremal orbit area is Onsager quantization. These measurements reveal the geometry and effective masses of a Fermi surface.

In a two-dimensional electron system, filling an integer number $\nu$ of resolved levels leads to the integer quantum Hall effect, withand very small longitudinal resistivity on a plateau. Disorder localization helps produce finite plateau widths. Cyclotron resonance provides another direct signature: radiation is absorbed at the adjacent-level spacing $\hslash {\omega }_c$.

Use the standard canonical anticommutation relations for the two distinct fermionic orbitals. In particular $\{a,b^{†}\}=\{a^{†},b\}=0$, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital fermionic Fock space interpretation. Normalize the original vacuum state by $⟨0|0⟩=1$ and fix the orderingThe canonical anticommutation relations give $a^2=(a^{†}{)}^2=0$ and $a{a}^{†}=1-a^{†}a$. Consequently the number operator is an orthogonal projector:The same holds for $N_b$. Hence each eigenvalue obeys $n^2=n$ and is either $0$ or $1$, expressing the Pauli exclusion principle. Also $[N_a,N_b]=0$, so the two number operators have a simultaneous orthonormal basis of Fock states:Each number operator has eigenvalues $0,1$, each with state degeneracy two. For $N_a$, the zero eigenspace is spanned by $|00⟩,|01⟩$ and the unit eigenspace by $|10⟩,|11⟩$; for $N_b$, these are respectively $|00⟩,|10⟩$ and $|01⟩,|11⟩$. Any normalized linear combination within one such eigenspace is also an eigenstate of that number operator.

The ordering convention will matter below: $b^{†}{a}^{†}|0⟩=-|11⟩$ and

Let $N_c=c^{†}c$, $N_d=d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Substitution of the Bogoliubov transformation, with normal ordering using $d{d}^{†}=1-d^{†}d$ and $c{c}^{†}=1-c^{†}c$, givesFor example, the $v^{2}d{d}^{†}$ term in $a^{†}a$ contributes $v^2-v^2{N}_d$; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequentlyTo eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfyingWith $u=\cos \theta$, $v=\sin \theta$, take $0<\theta <\pi /4$, so $\tan 2\theta =\Delta /\epsilon$. Setting $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$ givesThis positive choice fixes a convenient convention for the vacuum state in part (d). It yieldsBoth number operators independently have eigenvalues $0,1$. Hence the energy eigenvalues and state degeneracies areThere is one fermionic Bogoliubov quasiparticle of energy $E$ per occupied transformed orbital. The two Nambu spinor matrix eigenvalues $\pm E$ should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis $|0⟩,|11⟩$, the even-fermion parity block isbecause $ab|11⟩=-|0⟩$ and $b^{†}{a}^{†}|0⟩=-|11⟩$. Its characteristic polynomial is $x^2-2\epsilon x-{\Delta }^2$, giving $x=\epsilon \pm E$. The odd-fermion parity Fock states $a^{†}|0⟩,b^{†}|0⟩$ are unaffected by pairing and both have energy $\epsilon$.

Equivalently, with the Nambu spinor $\Psi =(a,b^{†}{)}^{\mathsf{T}}$,Replacing $\Psi$ by $R(c,d^{†}{)}^{\mathsf{T}}$ gives $E{c}^{†}c-Ed{d}^{†}+\epsilon =E(N_c+N_d)+\epsilon -E$, reproducing the result by normal ordering.

Use the Einstein coefficients convention in which $B_{ij}$ multiplies spectral radiation energy density $u_{\nu }$, with $u_{\nu }=c^{-1}\int I_{\nu }d\Omega$. A normalized spectral line profile obeys $\int \phi (\nu )d\nu =1$. The spontaneous emission probability per upper-state atom per unit time is $A_{21}$; the stimulated emission and upward absorption rates per atom are respectively $B_{21}\int u_{\nu }\phi (\nu )d\nu$ and $B_{12}\int u_{\nu }\phi (\nu )d\nu$.

For isotropic spontaneous emission, $n_2{A}_{21}$ transitions occur per unit volume and time, each supplying photon energy $h{\nu }_{12}$. Dividing among $4\pi$ steradians givesFor a small incident beam, its contribution to $u_{\nu }$ is $I_{\nu }d\Omega /c$. Absorption removes and stimulated emission adds photons in the incident direction. Matching the net loss to ${\alpha }_{\nu }{I}_{\nu }d\Omega$ therefore givesReplacing ${\nu }_{12}$ by $\nu$ across the narrow spectral line profile gives the printed expressions. The negative sign is essential: an inverted population can amplify the incident specific intensity. Thermal equilibrium and the Planck function imply $g_1{B}_{12}=g_2{B}_{21}$ and $A_{21}/B_{21}=8\pi h{\nu }_{12}^3/c^3$, where $g_i$ is the state degeneracy. A convention using mean intensity instead changes both $B$ coefficients by $4\pi /c$.

For the unresolved Lyman-alpha line, the lower $1s$ and combined upper $2p$ state degeneracies are $g_1=2$ and $g_2=6$, so the Einstein coefficients giveThe measured equivalent width $W=7.0\times 10^{-12}\mathrm{m}$ then impliesAssume a central transverse line of sight through a uniform filament, so $\ell =d=0.5\mathrm{Mpc}=1.54\times 10^{22}\mathrm{m}$. Nearly all neutral hydrogen is in its ground state at this temperature, henceThe cylinder's length $L$ determines its total volume but does not enter a transverse column density. Geometry must be specified: an axial line of sight has $\ell =L$ and gives $n_{\mathrm{HI}}=4.2\times 10^{-7}{\mathrm{m}}^{-3}$ for the same measured equivalent width. Intermediate chords give intermediate path-dependent estimates. The numbers above treat $W$ as a rest-frame equivalent width; an observed value at redshift $z$ must first be divided by $1+z$.

Using $dx/dT=-x/T$ in the internal energy givesThis is a Schottky anomaly with unequal degeneracies, with lower-level state degeneracy two and upper-level state degeneracy one. Its limiting forms areIt vanishes at low temperature because exciting the upper orientation is exponentially unlikely. It also vanishes at high temperature because the populations have nearly saturated and further heating hardly changes the internal energy. Between these limits it has one smooth maximum. The original plot shows its shape; numerical maximization gives $k_{\mathrm{B}}{T}_{\max }/\delta \simeq 0.449$ and $C_{\max }/(N{k}_{\mathrm{B}})\simeq 0.241$.

A planar rigid rotor has angular eigenfunctions $e^{im\varphi }$ and single-valuedness imposes $m=0,\pm 1,\pm 2,\dots$. Its energy depends on $m^2$, so if the level is labelled by $J=|m|$, thenThe two states at $J>0$ represent opposite directions of angular momentum. If the signed quantum number itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.

For an unrestricted three-dimensional linear rigid rotor, $J=0,1,2,\dots$ has $M=-J,\dots ,J$ and state degeneracy $2J+1$, with $E_J={\hslash }^{2}J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before nuclear spin statistics. For homonuclear hydrogen, the allowed even/odd spatial states must also be paired with the appropriate parahydrogen or orthohydrogen nuclear states, as considered explicitly in A8.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

Choose the ground-state energy as zero and write $x=\epsilon /(k_{B}T)$. The molecular partition function must sum individual states, so the triply degenerate excited level contributes three Boltzmann factors:The probabilities of occupying the two levels areEach of the three individual excited states has probability $e^{-x}/q$. At high temperature the four states become equiprobable, so the two levels have probabilities $1/4$ and $3/4$, not $1/2$ each. This is the distinction between a level and its state degeneracy.

Use the additive-volume two-level system implicit in the model: each excited particle increases the volume by the same increment. The difference between the two endpoint volumes is $V_0$, so that increment is $v=V_0/N$.

If $N_1$ particles are excited, $V=V_0+(V_0/N)N_1$. Linearity of the expectation gives $⟨N_1⟩=N{P}_1$, henceAs $T\to 0$, $V\to V_0$. As $T\to \infty$, only three quarters of the particles are excited on average, giving $V\to 7V_0/4$, rather than the all-excited endpoint $2V_0$. These limits also check the inclusion of state degeneracy.

For the one-dimensional freely jointed chain, counting right- and left-pointing links givesThereforeChoosing which links point right gives the state degeneracyPut $x=L/(Na)$. Applying the Stirling formula to $S=k_B\log \Omega$ and discarding terms subleading in $N$ yields

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.